Schur Index Divisibility Question: Ind A^E divides Ind A Background notation: If $A\in\mathcal{F}$ is a central simple algebra, $A\cong M_n(D)$, where $D$ is a division algebra. The Schur index of $A$ is defined as $Ind(A)=Deg(D)$. **How do we prove $A^E$ divides $Ind(A)$?** (We may assume further assume that $A$ is a division algebra). $A^E:=A\otimes E$, where $E$ is an extension field of $F$. According to Richard Pierce's book, it is supposed to follow easily from the following result (ii): $Ind(A)$ divides $Deg(A)$, and $Ind(A)=Deg(A)$ if and only if $A$ is a division algebra. My workings: I managed to prove (ii) as follows: $Deg(A)=nDeg(D)=nInd(A)$, so $Ind(A)\mid Deg(A)$. $Ind(A)=Deg(A)$, iff $n=1$, so $A\cong D$. Then, $$ Ind(A^E)\mid Deg(A\otimes E) $$ and I am stuck here. Can I claim that $A\otimes E\cong A$, and thus finish it off?

If I correctly recall the definitions, then

1. $\dim_FD=(\deg D)^2$ (this holds for all division algebras that are f.d. over their center).
2. $\dim_E A^E=\dim_F A$ (this is obvious).

3. If $D$ is a division algebra with center $F$, then $D\otimes_F E$ is a central simple algebra with center $E$. Therefore, by Wedderburn's Theorem, $$D\otimes_FE\simeq M_k(D_E)$$ for some division algebra $D_E$ with center $E$.

4. So $$k^2\dim_E D_E= \dim_E M_k(D_E)=\dim_E(D\otimes_FE)=\dim_F D.$$

5. Combining items 1 and 4 gives us $$(\deg D)^2=\dim_F D=k^2\dim_E D_E=k^2(\deg D_E)^2,$$ and consequently $$\deg D= k\cdot \deg D_E.$$

6. So $\deg D_E\mid \deg D$.




By your definition $\operatorname{Ind}(A)=\deg D$ and $\operatorname{Ind}(A^E)=\deg D_E$, so we are done.